# Are yields for organic farming different from conventional farming yields? Independent random samples from method A (organic farming) and method B (co

Are yields for organic farming different from conventional farming yields? Independent random samples from method A (organic farming) and method B (conventional farming) gave the following information about yield of sweet corn (in tons/acre). $\text{Method}A:6.51,7.02,6.81,7.27,6.73,6.11,6.17,5.88,6.69,7.12,5.74,6.90.$
$\text{Method}B:7.32,7.01,6.66,6.85,5.78,6.48,5.95,6.31,6.50,5.93,6.68.$ Use a 5% level of significance to test the claim that there is no difference between the yield distributions. (a) What is the level of significance? (b) Compute the sample test statistic. (Round your answer to two decimal places.) (c) Find the P-value of the sample test statistic. (Round your answer to four decimal places.)
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From the given data: ${\stackrel{―}{x}}_{1}=\frac{\sum {x}_{1}}{{n}_{1}}=\frac{78.95}{12}=6.579$
${s}_{1}=0.4992$
${\stackrel{―}{x}}_{2}={\stackrel{―}{x}}_{1}=\frac{\sum {x}_{2}}{{n}_{2}}=\frac{71.47}{11}=6.497$
${s}_{2}=0.4791$ Calculating pooled standard deviation: ${s}_{p}=\sqrt{\frac{{s}_{1}^{2}\left({n}_{1}-1\right)+{s}_{2}^{2}\left({n}_{2}-1\right)}{{n}_{1}+{n}_{2}-2}}=\sqrt{\frac{{0.4992}^{2}×11+{0.4791}^{2}×10}{12+11-2}=0.4897}$
$H0:{\mu }_{1}={\mu }_{2}$
$HA:{\mu }_{1}\ne {\mu }_{2}$

a) Leve of significance=0.05

b) Test Statistics: $t=\frac{{\stackrel{―}{x}}_{1}-{\stackrel{―}{x}}_{2}}{{s}_{p}×\sqrt{\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}}}}=\frac{6.579-6.497}{0.4897×\sqrt{\frac{1}{12}+\frac{1}{11}}}=0.40$

c)P-value for a two tailed test: Degree of freedom $={n}_{1}+{n}_{2}-2=12+11-2=21$ Using t-distribution table P-value $=0.6932$ As P-value is more than the significance level, we do not have sufficient evidence to reject H0 Conclusion: There is no difference between the yield distributions.